| Topic: |
Science > Physics |
| User: |
"GHooper" |
| Date: |
18 Jan 2008 06:01:51 PM |
| Object: |
Temperature drop across dissimilar insulators ? |
Short version.....
Is there a simple way to calculate
temperature drop across dissimilar insulators ?
Longer version....
I assume that for something like a wall thats
5 1/2 " with only one material like fiberglass
that the temperature drop is linier.
Fiberglass is about an R 3 per inch...
But put two different materials in the assembly,
two different R per inch values and the fun begins.
Say a simplified wall could be built like this
ambient @20 degreesF - 1" foam (R5/inch ) - 4.5" Fiberglass( R3/inch) -
sheetrock (R1) Conditioned space 72 deg F
What would the temperature be in between the foam and fiberglass ?
How did you perform the calculation ?
A question that I have not been able to solve ...
Thanks !
Glenn
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| User: "tadchem" |
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| Title: Re: Temperature drop across dissimilar insulators ? |
18 Jan 2008 06:55:45 PM |
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On Jan 18, 7:01 pm, "GHooper" <Hoop...@optoffline.net> wrote:
Short version.....
Is there a simple way to calculate
temperature drop across dissimilar insulators ?
Measure the temperature on each side with a thermometer.
Subtract the smaller reading from the larger reading.
The difference is the temperature drop.
Longer version....
I assume that for something like a wall thats
5 1/2 " with only one material like fiberglass
that the temperature drop is linier.
Fiberglass is about an R 3 per inch...
But put two different materials in the assembly,
two different R per inch values and the fun begins.
Say a simplified wall could be built like this
ambient @20 degreesF - 1" foam (R5/inch ) - 4.5" Fiberglass( R3/inch) -
sheetrock (R1) Conditioned space 72 deg F
The R-value is related to the thermal conductivity - that is, the RATE
at which heat flows.
http://en.wikipedia.org/wiki/R-value_(insulation)
Treat the temperatures as voltages and the R-values as resistances.
The flow of heat will act like an electric current.
What would the temperature be in between the foam and fiberglass ?
How did you perform the calculation ?
DIAGRAM:
72=B0 <-- R =3D 1 (sheetrock) --> T1 <-- R =3D 13.5 (3 * 4.5, fiberglass) --=
T2 <-- R =3D 5 ( 1 * 5, foam) --> 20=B0F
The temperature difference is 52=B0 across a total R of 19.5, so the
temperature drop is 2.67=B0 per unit R.
T1 =3D 72=B0 - (2.67=B0/R * 1R) =3D 69.33=B0
T2 =3D 69.33=B0 - (2.67=B0/R * 13.5R) =3D 33.275=B0
As a check,
T2 - 20=B0 =3D 2.67=B0/R * 5R, so T2 =3D 33.35=B0, which agrees to within th=
e
accuracy of the temperatures measured outside the walls.
A question that I have not been able to solve ...
HTH
Thanks !
Glenn
Tom Davidson
Richmond, VA
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| User: "GHooper" |
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| Title: Re: Temperature drop across dissimilar insulators ? |
18 Jan 2008 11:19:10 PM |
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Tadchem,
the purpose of calculating is for prediction, thats hardly any fun after the
fact :)
But dang, I almost didnt see your reply down lower !
Ultimatly, add the R values, then divide the delta T by total R to get
a per inch value....
Ahhh ....
Now I understand it visualy.
Gota love this stuff.....
Thanks
Glenn
"tadchem" <tadchem@comcast.net> wrote in message
news:0d900e1b-d07f-4028-ad18-bc7d4f723c0c@m34g2000hsb.googlegroups.com...
On Jan 18, 7:01 pm, "GHooper" <Hoop...@optoffline.net> wrote:
Short version.....
Is there a simple way to calculate
temperature drop across dissimilar insulators ?
Measure the temperature on each side with a thermometer.
Subtract the smaller reading from the larger reading.
The difference is the temperature drop.
Longer version....
I assume that for something like a wall thats
5 1/2 " with only one material like fiberglass
that the temperature drop is linier.
Fiberglass is about an R 3 per inch...
But put two different materials in the assembly,
two different R per inch values and the fun begins.
Say a simplified wall could be built like this
ambient @20 degreesF - 1" foam (R5/inch ) - 4.5" Fiberglass( R3/inch) -
sheetrock (R1) Conditioned space 72 deg F
The R-value is related to the thermal conductivity - that is, the RATE
at which heat flows.
http://en.wikipedia.org/wiki/R-value_(insulation)
Treat the temperatures as voltages and the R-values as resistances.
The flow of heat will act like an electric current.
What would the temperature be in between the foam and fiberglass ?
How did you perform the calculation ?
DIAGRAM:
72° <-- R = 1 (sheetrock) --> T1 <-- R = 13.5 (3 * 4.5, fiberglass) --
T2 <-- R = 5 ( 1 * 5, foam) --> 20°F
The temperature difference is 52° across a total R of 19.5, so the
temperature drop is 2.67° per unit R.
T1 = 72° - (2.67°/R * 1R) = 69.33°
T2 = 69.33° - (2.67°/R * 13.5R) = 33.275°
As a check,
T2 - 20° = 2.67°/R * 5R, so T2 = 33.35°, which agrees to within the
accuracy of the temperatures measured outside the walls.
A question that I have not been able to solve ...
HTH
Thanks !
Glenn
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: Temperature drop across dissimilar insulators ? |
18 Jan 2008 06:38:28 PM |
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Yep, this is a fairly simple calculation. Temperature is linear within
each material as you suspect.
First find the thermal power transmitted through the 3 layers:
P = total temperature difference / total thermal resistance
P = (72F - 20F) / ( R5 * 1" + R3*4.5" + R1 )
Then use this to find the temperature at any boundary, eg:
T_foam-fibreglass = 20F + P * R5*1"
If you're familiar with electric circuits and ohm's law, this is easy
to visualise using an equivalent circuit. See
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics
I assume R5 and R3 are thermal resistance per unit thickness. And that
they include the finite area of the sheets. If they're really times
unit area then the calculation is the same but power should be called
power/unit area.
On Jan 19, 1:01 pm, "GHooper" <Hoop...@optoffline.net> wrote:
ambient @20 degreesF - 1" foam (R5/inch ) - 4.5" Fiberglass( R3/inch) -
sheetrock (R1) Conditioned space 72 deg F
What would the temperature be in between the foam and fiberglass ?
How did you perform the calculation ?
A question that I have not been able to solve ...
Thanks !
Glenn
.
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| User: "GHooper" |
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| Title: Re: Temperature drop across dissimilar insulators ? |
18 Jan 2008 11:08:42 PM |
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Too cool !
Even as I wrote the problem it reminded me of two resistors in a series
circut. But I couldnt think that last equasion through ..
So according to this, you think it should be 33.6 Deg ?
If so this is most excellent !!!!
Glenn
<nottoooily@hotmail.com> wrote in message
news:df41854a-1bbe-4088-a3a7-b0e68fdfe065@s12g2000prg.googlegroups.com...
Yep, this is a fairly simple calculation. Temperature is linear within
each material as you suspect.
First find the thermal power transmitted through the 3 layers:
P = total temperature difference / total thermal resistance
P = (72F - 20F) / ( R5 * 1" + R3*4.5" + R1 )
Then use this to find the temperature at any boundary, eg:
T_foam-fibreglass = 20F + P * R5*1"
If you're familiar with electric circuits and ohm's law, this is easy
to visualise using an equivalent circuit. See
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics
I assume R5 and R3 are thermal resistance per unit thickness. And that
they include the finite area of the sheets. If they're really times
unit area then the calculation is the same but power should be called
power/unit area.
On Jan 19, 1:01 pm, "GHooper" <Hoop...@optoffline.net> wrote:
ambient @20 degreesF - 1" foam (R5/inch ) - 4.5" Fiberglass( R3/inch) -
sheetrock (R1) Conditioned space 72 deg F
What would the temperature be in between the foam and fiberglass ?
How did you perform the calculation ?
A question that I have not been able to solve ...
Thanks !
Glenn
.
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